When ballots are complete rankings, margins and winning votes give the same order of defeat strength. This is easy to see because for complete rankings the entries of the pairwise matrix satisfy P(j, i) = N - P(i, j), where N is the total number of ballots ... because on each ballot where candidate i is not ranked ahead of candidate j, candidate j is ranked of candidate i. So if candidate i defeats candidate j pairwise, then the winning votes are wv = P(i, j) = N - P(j, i), and the defeat margin is P(i, j) - P(j, i) which is also equal to P(I, j) - (N - P(i, j)), which simplifies to margin = 2*P(I, j) - N, or margin = 2*wv - N This relation is order preserving: (margin1 > margin2) iff (wv1 > wv2) In other words, when ballot rankings are complete, wv and margins give the same order to pairwise defeat strengths. Another basic fact is that symmetric completion does not change pairwise margins ... adding 1/2 to both P(i, j) and P(j, i) does not change their difference. So using margins to measure defeat strength is equivalent to using margins after symmetric completion, which is equivalent to using wv after symmetric completion, because for complete (or completed) rankings wv and margins give the same order of defeat strengths. Now, it is clear that since Chicken attacks by definition are carried out by truncations, when truncations are not allowed there can be no chicken defection. One might be tempted to conclude that automatic symmetric completion would repair or thwart a chicken attack, but this is not quite right because a truncation followed by a symmetric completion only repairs half the truncation and turns the other half into a burial, which could very well be an unintended consequence. This is one reason why a check box for symmetric completion is not such a good idea for chicken attack resistance, after all. I hope this sheds further light on the relationships among winning votes, margins, symmetric completions, and Chicken defense. Forest